Solving an elliptic system from compatible boundary measurements

An elliptic equation leaves a finite-dimensional family of decaying normal modes at each nonzero boundary frequency. Boundary measurements must distinguish exactly that family. We first solve the resulting projected operator problem, then prove the estimates that turn its algebraic inverse into an inverse on Sobolev spaces. The last step identifies every finite-dimensional obstruction as a smooth density-valued section, including at the boundary.

The interior operator acts between arbitrary complex bundles of equal finite rank. Its order is any positive integer, and the total orders of the boundary measurements may exceed it. Neither self-adjointness, positivity, scalar coefficients, nor a Dirichlet boundary condition is assumed. All assertions use restriction Sobolev spaces on a compact smooth manifold with boundary. The boundary realization below is taken at $s\geq m$; a separate potential estimate will hold at every real $s$.

1. The spaces, their weights, and the exact interfaces

Let $X$ be a compact smooth manifold with boundary $Y$, and let $E,F$ be smooth complex bundles of common finite rank. Choose a collar $(y,t)$, with $t\geq0$ pointing into $X$, and identify its bundles with pullbacks. Write $D=-i\partial$. An elliptic differential operator $P:E\to F$ of order $m\geq1$ has, in the collar, the form $$P=\sum_{a=0}^m P_a(y,t,D_y)D_t^a,\qquad \operatorname{ord}P_a\leq m-a,\qquad P_m\text{ invertible}. \tag{BF1}$$ Boundary operators of transversal order below $m$ have the form $$B_j u=\sum_{k=0}^{m-1}\mathcal B_{jk}\gamma_k u, \quad\gamma_k u=(D_t^ku)|_{t=0},\quad \operatorname{ord}\mathcal B_{jk}\leq m_j-k. \tag{BF2}$$ Here $G_j$ is the target bundle, and zero entries are allowed. Initially the $\mathcal B_{jk}$ are differential operators. Every proof also permits classical tangential pseudodifferential entries of the displayed degrees. In that extension the $m_j$ may be real. The interior operator remains differential. We do not infer the whole boundary calculus for arbitrary pseudodifferential interior operators from this extension.

Put $\mathcal E=E|_Y^{\oplus m}$, $\mathcal G=\bigoplus_jG_j$, and $$\mathcal C^s=\bigoplus_{k=0}^{m-1}H^{s-k-1/2}(Y,E),\quad \mathcal D^s=\bigoplus_jH^{s-m_j-1/2}(Y,G_j),\quad \mathcal Y^s=\bar H^{s-m}(X^\circ,F)\oplus\mathcal D^s. \tag{BF3}$$ The bar denotes the restrictions of whole-manifold Sobolev distributions, with quotient norm; it does not impose zero boundary jets. Our realization is $A_s:\bar H^s(X^\circ,E)\to\mathcal Y^s$, $A_su=(Pu,Bu)$, for $s\geq m$. The trace theorem gives $\gamma:\bar H^s\to\mathcal C^s$, since $s>m-1+1/2$. Applying each tangential entry in (BF2) then proves continuity of $A_s$, including negative boundary target orders when an $m_j$ is large.

The proof uses the restriction and extension Sobolev scale, Fourier inversion, the trace theorem at $s>k+1/2$, smooth multiplication, partitions of unity, compact Sobolev inclusions and interpolation of restriction spaces. The Euclidean symbol calculus and coordinate-free microlocal calculus supply asymptotic summation, composition, formal transpose, Sobolev mapping and interior elliptic regularity. Singularities along a submanifold and smooth boundary passage, Sections 10–13, supplies ordinary transmission and the layer traces. Cauchy data from jumps and residues, Sections 6–12, supplies the jump, projection and smoothing interfaces displayed below. We use Hilbert completeness, Cauchy–Schwarz, bounded adjoints and orthogonal projections; the required functional representation is proved in Section 2. Fredholm stability, Sections 3 and 5, supplies norm stability, compact perturbations and the two-parametrix criterion. Each earlier result is used with its stated hypotheses.

Here is the Calderón interface with signs fixed. Extend $P$ elliptically across $Y$, and take a proper classical transmission parametrix $T$ with $TP=I+R_E$, $PT=I+R_F$, where both remainders have smooth kernels. Define zero extension $e^+$, interior restriction $r^+$, and $$V=r^+Te^+,\qquad K=r^+TP^c,\qquad P^cU=\frac1i\sum_{a=1}^m P_a\sum_{r=0}^{a-1} U_{a-1-r}\otimes D_t^r\delta_0. \tag{BF4}$$ For smooth inputs, the interface asserts $$u+R_E^+u=VPu+K\gamma u,\quad \gamma K=Q,\quad Q_{kl}\in\Psi^{k-l}_{\rm cl},\quad Q^2-Q\in\Psi^{-\infty},\qquad R_E^+=r^+R_Ee^+. \tag{BF5}$$ Moreover $Q\gamma V:L^2(X,F)\to C^\infty(Y,\mathcal E)$ is continuous. Notice which projection occurs here. The complementary projection applied to $\gamma V$ need not be smoothing. The principal weighted symbol $q$ projects onto all Cauchy jets of decaying solutions of the frozen normal equation; its complementary space consists of the opposite normal modes. Smoothness and multiplicities of those spaces are the matrix spectral assertions in Sections 1–4 of Stable modes and the algebra of boundary data, without an eigenbasis assumption.

The original principal projection has the exact antipodal correspondence proved in Cauchy data from jumps and residues, Section9. Write $N=\operatorname{rank}E=\operatorname{rank}F$. In $\dim X\geq3$, its rank is necessarily $mN/2$, and the complementing bijection therefore forces $\operatorname{rank}\mathcal G=\sum_j\operatorname{rank}G_j=mN/2$. This conclusion also applies to the real-order tangential boundary extension here because the interior principal polynomial remains differential and homogeneous. Equal target rank alone does not prove complementing; the explicit three-dimensional scalar example in that proof has equal rank and a zero measurement at a nonzero covector. In dimension two a fixed differential target can be complementing on both rays only if their complementary stable ranks agree, while interior ellipticity alone need not give that agreement. No such tangential assertion is imposed on dimension one.

All smoothing statements on a compact boundary mean a smooth kernel. When one input is an interior $L^2$ function we specify that domain separately; a continuous map from $L^2$ into $C^\infty$ is not being declared a smooth-kernel operator on arbitrary interior distributions.

2. Constructing order reductions without an index assumption

Let $H$ be a smooth Hermitian bundle on a closed compact manifold. For each real $a$ there is an invertible classical operator $$J_H^a:H^r(Y,H)\longrightarrow H^{r-a}(Y,H) \quad\text{for every real }r, \tag{BF6}$$ with positive scalar principal symbol $|\eta|^a I_H$. Its inverse is classical of order $-a$. We need actual inverses so that changing weights adds no undisclosed Fredholm index.

2.1. The earlier construction and its exact defect

Editorial correction. The following earlier construction is retained for comparison. Its claimed classical membership and classical inverse fail at a positive noninteger order. The full norm and mapping estimates do not remove that degree-zero defect. The corrected construction in Section2.2 proves (BF6) at every original real exponent.

For $a=2b>0$, choose an elliptic operator $C$ of order $b$ with principal symbol $|\eta|^b I_H$. This is obtained by patching that symbol and quantizing. The operator $D=I+C^*C$ has order $2b$, is formally self-adjoint and has the required positive principal symbol. The elliptic parametrix gives $\|u\|_{H^b}\leq C_1(\|Cu\|_{L^2}+\|u\|_{L^2})$: apply a parametrix of $C$, and bound its smoothing remainder from $L^2$ to $H^b$. Thus the quadratic form $\|u\|_{L^2}^2+\|Cu\|_{L^2}^2$ defines an equivalent Hilbert norm on $H^b$. The Hilbert representation theorem applied to that inner product solves $Du=f$ uniquely for $f\in H^{-b}$. For $r\geq b$ and $f\in H^{r-2b}$, elliptic regularity upgrades the same solution to $H^r$, and the parametrix estimate proves boundedness there. For $r<b$, transpose the bounded inverse at exponent $2b-r>b$; self-adjointness yields a bounded inverse from $H^{r-2b}$ to $H^r$, with both inverse identities extending from smooth sections by density. Uniqueness follows from the form identity after elliptic regularity, including for distributional nullvectors. A two-sided parametrix $D_0$ satisfies $D^{-1}-D_0=-D^{-1}(DD_0-I)$; this remainder has a smooth kernel because elliptic regularity applies also to the transposed expression. Hence $D^{-1}\in\Psi^{-2b}_{\rm cl}$. Set $J_H^a=D$. For $a<0$, invert the construction for $-a$; set $J_H^0=I$. No group law between separately chosen $J_H^a$’s is needed.

Here is the functional representation used in the form step. For a nonzero bounded functional $f$ on a Hilbert space, its kernel is closed. Choose $z$ outside that kernel and subtract its orthogonal projection onto the kernel, obtaining $z_0\ne0$ orthogonal to it. For every $u$, the vector $u-f(u)z_0/f(z_0)$ lies in the kernel. Pairing it with $z_0$ expresses $f(u)$ as a fixed inner product with $u$, with the conjugations determined by the chosen convention. The zero functional is represented by zero. This proves the needed representation from the Hilbert facts stated in Section 1. Apply it to the equivalent form inner product on $H^b$; it is complete because its norm is equivalent to the Sobolev norm. This is not an invocation of a boundary solvability theorem.

2.2. The corrected operator, its inverse, and every original contribution

Keep the original closed compact smooth manifold $Y$, Hermitian bundle $H$, density, cotangent norm, every real exponent and step-one classical expansion. The receiving calculus, elliptic regularity, compact Sobolev inclusion and Hilbert representation are the existing in-course providers. The original $D=I+C^*C$ is retained for comparison. The corrected operator is separately named $D_K$.

1. The obstruction for every noninteger positive order

Let $a>0$ be noninteger, let $q=\lfloor a\rfloor$, and suppose $Y$ has positive dimension and $H$ has nonzero rank on the component under consideration. For $P\in\Psi^a_{\mathrm{cl}}(Y;H)$, its local symbol has the step-one expansion $p\sim\sum_{j\geq0}p_{a-j}$, with original homogeneous coefficients. The first $q+1$ coefficients of any putative classical expansion of $I+P$ must be these same coefficients: successively divide their difference by each positive degree $|\eta|^{a-j}$ and take the ray limit. The identity has strictly smaller degree than each of them. Uniqueness of homogeneous coefficients therefore gives $$\sigma(I+P)-\sum_{j=0}^{q}p_{a-j} =I_H+r,\qquad r\in S^{a-q-1},\qquad a-q-1<0. \tag{OR1}$$ On each fixed nonzero ray $r(y,t\eta)\to0$, whereas the left side tends to $I_H\ne0$. It cannot belong to $S^{a-q-1}$, as a classical order-$a$ remainder would. Thus the obstruction applies to every such noninteger $a$, not only one example. It is the additional degree-zero identity coefficient, lying outside $\{a-j:j\in\mathbb N_0\}$. The original full symbol still belongs to the ordinary order-$a$ symbol class; all its terms remain. For positive integer $a$, degree zero is already in that list and the identity can be added to that exact coefficient. No alteration of the homogeneous step is used.

For the explicit circle example, use $Y=\mathbb R/(2\pi\mathbb Z)$, density $dx$, scalar bundle and basis $e_k(x)=(2\pi)^{-1/2}e^{ikx}$. The self-adjoint elliptic multiplier $$Ce_k=(1+k^2)^{1/8}e_k,\qquad b=\tfrac14, \qquad De_k=[1+(1+k^2)^{1/4}]e_k, \quad a=2b=\tfrac12 \tag{OR2}$$ retains the full original $1+k^2$. Its classical status follows directly from the finite Taylor formula for $(1+z)^{1/8}$: for each positive integer $L$, its coefficients are $\binom{1/8}{l}$, $0\leq l<L$, and its integral remainder is $$\frac{z^L}{(L-1)!}\int_0^1(1-t)^{L-1} \left[\prod_{h=0}^{L-1}(\tfrac18-h)\right] (1+tz)^{1/8-L}\,dt . \tag{OR3}$$ Put $z=\xi^{-2}$ and multiply by $|\xi|^{1/4}$ on both frequency rays. Every odd step-one homogeneous coefficient is zero; every even coefficient and all derivatives of the integral remainder are retained. The smooth full symbol at small frequency is $(1+\xi^2)^{1/8}$. The forced leading symbol of $D$ is $|\xi|^{1/2}$, but $$1+(1+\xi^2)^{1/4}-|\xi|^{1/2}\longrightarrow1. \tag{OR4}$$ Taylor’s formula at zero gives $(1+z)^{1/4}-1=O(z)$, so the extra difference without the identity is $O(|\xi|^{-3/2})$; it cannot cancel the one. The classical first remainder would have order $-1/2$. This contradiction holds also on the integer frequency sequence defining the circle multiplier.

The inverse has the same exact defect. With $t=|\xi|$ and $d(\xi)=1+(1+\xi^2)^{1/4}$, $$d(\xi)^{-1}-t^{-1/2} =\frac{t^{1/2}-d(\xi)}{t^{1/2}d(\xi)}, \qquad t[d(\xi)^{-1}-t^{-1/2}]\longrightarrow-1. \tag{OR5}$$ An order-$-1/2$ classical inverse would have first remainder of order $-3/2$, which after multiplication by $t$ would tend to zero. Thus the inverse isomorphism does not establish the asserted classical inverse.

2. The exact finite-rank correction

For the original $a=2b>0$, keep the chosen classical elliptic $C\in\Psi^b_{\mathrm{cl}}(Y;H)$ with principal symbol $|\eta|^b I_H$. Its original parametrix gives, for a fixed finite $C_1$, $$\|u\|_{H^b}\leq C_1(\|Cu\|_{L^2}+\|u\|_{L^2}), \qquad u\in H^b(Y,H). \tag{OR6}$$ Let $K=\ker(C:H^b\to L^2)$. It is closed. A distributional kernel section is smooth: if $SC=I+R_C$ with smooth-kernel remainder, then $u=-R_Cu$ whenever $Cu=0$. The $H^b$ unit ball of $K$ is compact. Indeed compact $H^b\hookrightarrow L^2$, valid because $b>0$, gives an $L^2$-convergent subsequence; (OR6) applied to its differences upgrades it to an $H^b$-Cauchy subsequence in the closed space $K$.

For completeness this implies finite dimension. If a normed space were infinite dimensional, start with any finite-dimensional subspace $M$, which is closed. Choose $x\notin M$, write $\delta=\inf_{w\in M}\|x-w\|>0$, and choose $w\in M$ with $\|x-w\|<2\delta$. The unit vector $(x-w)/\|x-w\|$ has distance greater than $1/2$ from $M$. Repeating with the span of the previously chosen vectors gives a sequence in the unit ball separated by more than $1/2$, contradicting compactness. This proves the needed finite-dimensional claim without an index assumption on $C$.

Choose an $L^2$-orthonormal smooth basis $e_1,\ldots,e_d$ of $K$, using the fixed original metric and density. Write $(u,v)_{L^2}$ linear in $u$. The exact projection is $$\Pi_Ku=\sum_{j=1}^d(u,e_j)_{L^2}e_j, \qquad \Pi_K^*=\Pi_K,\quad\Pi_K^2=\Pi_K, \qquad D_K=C^*C+\Pi_K. \tag{OR7}$$ The empty sum is zero when $d=0$. Its full kernel is $\sum_je_j(y)\otimes e_j(y')^*$, integrated against the original density in $y'$; it is smooth. Each coefficient is distributional pairing with a fixed smooth section, so $\Pi_K$ is bounded between every pair of Sobolev spaces. In particular it is a smoothing operator, not a discarded degree-zero identity. Thus $D_K\in\Psi^{2b}_{\mathrm{cl}}$, with principal symbol $|\eta|^{2b}I_H$.

On $H^b$ define $$q_K(u,v)=(Cu,Cv)_{L^2}+(\Pi_Ku,\Pi_Kv)_{L^2}, \quad q_K(u,u)=\|Cu\|_{L^2}^2+\|\Pi_Ku\|_{L^2}^2. \tag{OR8}$$ The upper bound is $q_K(u,u)\leq(\|C\|_{H^b\to L^2}^2+\|\Pi_K\|_{H^b\to L^2}^2)\|u\|_{H^b}^2$. There is also $\gamma>0$ such that $q_K(u,u)\geq\gamma\|u\|_{H^b}^2$. Otherwise, for each positive integer $n$, choose a nonzero vector with form-to-norm-squared ratio less than $1/n$, and divide it by its exact $H^b$ norm to obtain $u_n$ of norm one. Then $Cu_n\to0$ and $\Pi_Ku_n\to0$ in $L^2$. Compact inclusion gives an $L^2$-Cauchy subsequence. Equation (OR6) applied to differences makes it Cauchy in $H^b$, with limit $u$ of norm one. Continuity gives $Cu=0$ and $\Pi_Ku=0$. But $u\in K$ entails $\Pi_Ku=u$, a contradiction. If the section space is zero, any positive $\gamma$ gives the vacuous estimate. Every norm and both original maps are retained in this argument.

3. The exact inverses on every Sobolev space

Formal self-adjointness and projection idempotence give $\langle D_Ku,v\rangle=q_K(u,v)$, first on smooth sections and then on $H^b$ by density. The pairing is linear in its distribution and conjugate linear in the test section. For $f\in H^{-b}$, the functional $v\mapsto\langle f,v\rangle$ is bounded in the complete equivalent form norm. The Hilbert representation proved in Section2 of the lesson therefore gives unique $u\in H^b$ satisfying this identity for all $v\in H^b$. Its exact estimate is $$\gamma\|u\|_{H^b}^2\leq q_K(u,u) =\langle f,u\rangle\leq\|f\|_{H^{-b}}\|u\|_{H^b}, \qquad \|u\|_{H^b}\leq\gamma^{-1}\|f\|_{H^{-b}}. \tag{OR9}$$ For complex pairings the middle scalar is real nonnegative by the form identity; its absolute value supplies the displayed inequality. These are actual distributional solutions, and testing all smooth sections gives $D_Ku=f$.

For $r\geq b$, $f\in H^{r-2b}$ also belongs to $H^{-b}$. Elliptic regularity and the parametrix imply $u\in H^r$ and a bounded inverse there. More explicitly $S_KD_K=I+R_K$ gives $u=S_Kf-R_Ku$; the mapping estimates yield $$\|u\|_{H^r}\leq\|S_K\|_{H^{r-2b}\to H^r}\|f\|_{H^{r-2b}} +\|R_K\|_{H^b\to H^r}\gamma^{-1} \|\iota\|_{H^{r-2b}\to H^{-b}}\|f\|_{H^{r-2b}}. \tag{OR10}$$ No smoothing contribution or embedding norm is removed.

For $r<b$, set $p=2b-r>b$. The adjoint of the already established inverse $T_p:H^{p-2b}\to H^p$ acts from $H^{-p}$ to $H^{2b-p}$, which are exactly $H^{r-2b}$ and $H^r$. Formal self-adjointness of $D_K$ transposes both inverse identities. They extend from smooth sections by their density and the bounded Sobolev maps. A distributional nullvector of $D_K$ is smooth by its parametrix and is zero by (OR8); hence all these inverses agree on their common domains. Denote the resulting compatible inverse by $T=D_K^{-1}$.

Let $D_0\in\Psi^{-2b}_{\mathrm{cl}}$ be a classical parametrix and retain its whole right error $R= D_KD_0-I$. Then $$T-D_0=-TR. \tag{OR11}$$ This is smoothing, with an explicit kernel argument. In a finite collection of bundle charts let $R(\cdot,y')$ be its smooth output-variable kernel. For every input derivative $\beta$ and every real $N$, $$\|\partial_{y'}^\beta T R(\cdot,y')\|_{H^N} \leq\|T\|_{H^{N-2b}\to H^N} \|\partial_{y'}^\beta R(\cdot,y')\|_{H^{N-2b}}. \tag{OR12}$$ The right side is continuous and locally uniformly bounded in $y'$. Difference quotients pass through the bounded map $T$, so these are its actual derivatives. Taking $N$ above every requested output derivative order plus $\dim Y/2$ and using Sobolev embedding proves all jointly continuous mixed kernel derivatives. All chart and density factors remain those of the original smooth kernel. Thus $T\in\Psi^{-2b}_{\mathrm{cl}}$. Set $J_H^a=D_K$ for $a>0$, $J_H^a=(J_H^{-a})^{-1}$ using a separately constructed positive-order operator for $a<0$, and $J_H^0=I$. These give precisely (BF6), with its full original spaces and principal symbols, without an index assumption or a group law.

If $Y$ is zero dimensional, compactness makes it a finite set and all bundle-section spaces finite dimensional. Take the actual identity for each $J_H^a$. All kernels are smooth and all operators belong to every smoothing/classical order; the nonzero-covector symbol condition has empty domain. The empty manifold and zero bundle also have their unique invertible zero-space identity maps.

4. Comparison with the original operator and all receiving maps

The original and corrected operators have the exact relation $$D-D_K=I-\Pi_K, \qquad \big[\|Cu\|_{L^2}^2+\|u\|_{L^2}^2\big]-q_K(u,u) =\|(I-\Pi_K)u\|_{L^2}^2. \tag{OR13}$$ The last identity uses the original orthogonal decomposition, including the kernel component. On $K$ both operators equal the identity. The corrected form remains coercive on the entire original space, including $K^\perp$, by (OR6)–(OR9). In the explicit circle example $K=0$, so $D_K=C^*C$ has full multiplier $(1+k^2)^{1/4}$, and its actual inverse has multiplier $(1+k^2)^{-1/4}$; both retain every endpoint factor and belong to the required classical classes. This proves the mathematical bridge to the prior construction while repairing its precise degree-zero defect.

For the original operator matrix entry orders $b_i-a_j$, retain the actual diagonal maps $D_a=\operatorname{diag}(J^{a_j})$, $D_b=\operatorname{diag}(J^{b_i})$. The corrected $J$’s prove each map $H^r\to H^{r-a_j}$ and $H^r\to H^{r-b_i}$. Therefore $D_b^{-1}MD_a$ is classical of order zero, and restoring variables recovers every original entry order. For the boundary system, retain $a_k=k$, $b_j=m_j$, $r=s-1/2$ and both conjugations in (BF11). The corrected operators preserve the scalar positive principal-symbol isomorphisms and all smoothing ideals. Thus the order-minus-one errors, projected series and (BF12)–(BF13) use the exact step-one calculus at every real $m_j$. The separate realization restriction $s\geq m$, all layer Fourier factors and every density-valued obstruction are unchanged. This establishes the receiving proof, rather than inferring it from a nonclassical isomorphism.

The general noninteger obstruction (OR1), the explicit symbol and inverse tests (OR2)–(OR5), and the full repaired operator (OR6)–(OR13) concern this course’s local construction. This calculation does not attribute the faulty construction to another author.

5. Every derivative of the full frequency remainder

Write the integral factor after $z^L$ in (OR3) as $H_L(z)$, with $L\geq1$. Its $p$-th derivative on $[0,1]$ is obtained by multiplying the integrand by $t^p\prod_{h=0}^{p-1}(1/8-L-h)$ and replacing the exponent by $1/8-L-p$; these derivatives are bounded because $1+tz\geq1$. After substituting $z=\xi^{-2}$, the full differentiated frequency remainder is $$\partial_\xi^j\big[|\xi|^{1/4}\xi^{-2L}H_L(\xi^{-2})\big] =\sum_{h=0}^j\binom jh\partial_\xi^h(|\xi|^{1/4}\xi^{-2L}) \sum_{\pi\in\mathcal P_{j-h}}H_L^{(|\pi|)}(\xi^{-2}) \prod_{B\in\pi}(-1)^{|B|}(|B|+1)!\xi^{-|B|-2}. \tag{OR3a}$$ Here $\mathcal P_l$ consists of all set partitions of $\{1,\ldots,l\}$; its empty partition at $l=0$ gives $H_L$. This formula follows by the full product rule and the repeated chain rule: a new derivative either creates a singleton block or joins exactly one existing block, which generates every partition once. The exact derivative $\partial_\xi^l\xi^{-2}=(-1)^l(l+1)!\xi^{-l-2}$ supplies every factor. On each frequency ray the first factor is the falling product $\prod_{v=0}^{h-1}(1/4-2L-v)$, times $\operatorname{sgn}(\xi)^h|\xi|^{1/4-2L-h}$. A partition contributes $|\xi|^{-(j-h)-2|\pi|}$. Thus each term is bounded by its complete finite coefficient times $|\xi|^{1/4-2L-j}$ for $|\xi|\geq1$. This proves every symbol derivative bound while retaining every zero coefficient and ray sign.

6. The full original inverse and its exact comparison

The original form remains coercive, with the explicit bound $$\|Cu\|_{L^2}^2+\|u\|_{L^2}^2 \geq(2C_1^2)^{-1}\|u\|_{H^b}^2. \tag{OR14}$$ It follows by squaring (OR6) and using $(x+y)^2\leq2(x^2+y^2)$. Hilbert representation gives the original inverse $T_D:H^{-b}\to H^b$. Its every-real-order extension can be proved without assigning false classical membership to $D$. A parametrix $S_A$ of the retained classical elliptic $A=C^*C$, with $S_AA=I+R_A$, gives the full identity $u=S_Af-S_Au-R_Au$ when $Du=f$. If $u\in H^s$, $r\geq s$ and $f\in H^{r-2b}$, it upgrades $u$ to $H^{\min(r,s+2b)}$. Starting with $s=b$ reaches every prescribed $r\geq b$ in finitely many steps since $2b>0$. Each step retains the bounded maps $S_A:H^{r-2b}\to H^r$, $S_A:H^s\to H^{s+2b}$, $R_A:H^s\to H^r$ and their embedding norms; their finite composition yields a bounded inverse at that level. Transposing at $2b-r$ gives the smaller exponents as in Section3. A distributional nullvector belongs to some Sobolev space on compact $Y$. The same parametrix bootstrap, with $f=0$, successively raises its exponent by $2b$, makes it smooth, and the positive original form then forces it to vanish. Thus the inverses are compatible.

Both actual inverses satisfy the exact comparison on every distributional Sobolev domain: $$T_D-T=-T_D(I-\Pi_K)T: H^{r-2b}\longrightarrow H^{r+2b},\qquad r\in\mathbb R. \tag{OR15}$$ Indeed applying $D$ gives $- (D-D_K)T$; apply its compatible inverse. The middle map acts on $H^r$, and $T_D$ at exponent $r+2b$ maps $H^r$ into $H^{r+2b}$, proving the whole stated gain. No identity term or projection component has been suppressed. On $K$ both inverses are the identity; the difference is zero there. The exact relation also retains the original nonkernel contribution and agrees with the full circle multipliers in (OR2)–(OR5). These are proved receiving consequences of the reconstruction, with no historical novelty claim.

2.3. Restoring every original entry degree

More generally suppose an operator matrix $M=(M_{ij})$ has entry orders $b_i-a_j$, so its domain factors have exponents $r-a_j$ and targets have exponents $r-b_i$. Put $$D_a=\operatorname{diag}(J^{a_j}),\qquad D_b=\operatorname{diag}(J^{b_i}). \quad\widehat M=D_b^{-1}MD_a\in\Psi^0_{\rm cl}. \tag{BF7}$$ Each diagonal map identifies the equal-exponent space $\bigoplus H^r$ with the weighted one. Restoring the original variables recovers each entry order individually. In particular weighted smoothing means every entry is smoothing and is preserved by these conjugations.

3. One-sided inversion on a projected bundle

Let $Q\in\Psi^0_{\rm cl}(Y;H,H)$ satisfy $Q^2-Q\in\Psi^{-\infty}$, and let $B\in\Psi^\mu_{\rm cl}(Y;H,G)$, with real $\mu$. Write $q,b$ for their principal symbols. The image of $q$ is a smooth bundle on $T^*Y\setminus0$: the rank of a smooth idempotent is locally constant, and a nonzero minor supplies local frames. This bundle need not be the pullback of a bundle on $Y$.

If $b:qH\to G$ is onto at every nonzero covector, there exists $S\in\Psi^{-\mu}_{\rm cl}(Y;G,H)$ such that $$BS\equiv I_G,\qquad QS\equiv S. \tag{BF8}$$ Throughout this item $\equiv$ denotes equality modulo smoothing. To prove it, first reduce $\mu$ to zero using (BF6). The symbol $c=bq:H\to G$ is onto. Give both bundles metrics. Then $cc^*$ is positive definite and $t_0=c^*(cc^*)^{-1}$ is a smooth homogeneous right inverse. Quantize it to $T_0$; the error $I-BQT_0$ has order $-1$. Asymptotically summing $T_0\sum_{l\geq0}(I-BQT_0)^l$ gives $T_1$ with $BQT_1\equiv I$, since after $N$ terms the error is the $N$-th power of an order $-1$ operator. The complete calculus makes the final error smoothing. Set $S=QT_1$. Then $BS\equiv I$ and $QS-S=(Q^2-Q)T_1$ is smoothing. Restoring weights gives order $-\mu$.

If instead $b:qH\to G$ is one-to-one, there are $S'\in\Psi^{-\mu}_{\rm cl}(Y;G,H)$ and $S''\in\Psi^0_{\rm cl}(Y;H,H)$ with $$S'B+S''\equiv I_H,\qquad S''Q\equiv0. \tag{BF9}$$ After the same order reduction, consider the column $C=(B,I-Q)^t$. If $(I-q)v=0$ and $bv=0$, then $v\in qH$ and injectivity gives $v=0$. Thus its symbol $c$ is injective; $(c^*c)^{-1}c^*$ is a smooth left inverse. Quantization and the left version of the preceding asymptotic series give a row $(T',T'')$ with $T'B+T''(I-Q)\equiv I$. Set $S'=T'$, $S''=T''(I-Q)$. Multiplying by $Q$ gives the second identity because $(I-Q)Q$ is smoothing. This proves both statements for arbitrary ranks. Neither assertion alone promises a two-sided inverse for the original boundary realization.

4. Uniqueness and restoration of every entry degree

When $b:qH\to G$ is bijective, choose the operators from both parts of Section 3. In the quotient algebra, $$S'=S'BS=(S'B+S'')S=S, \tag{BF10}$$ where the added term vanishes because $QS=S$ and $S''Q=0$. Thus the same $S$ works in both identities. Comparing any other permitted right inverse with this fixed left pair proves its uniqueness; comparing a new left pair with the fixed right inverse proves uniqueness of $S'$, and then $S''=I-SB$ is unique. This uniqueness is modulo smoothing, not equality of particular quantizations, and it fails in general if only one of the two symbol conditions holds.

Apply this to (BF2) and (BF5). The source weights are $a_k=k$, the target weights are $b_j=m_j$, and the common exponent is $r=s-1/2$. The conjugated operators are $$\widehat Q=D_a^{-1}QD_a,\qquad \widehat{\mathcal B}=D_b^{-1}\mathcal B D_a. \tag{BF11}$$ They have order zero. The complementing condition is precisely that the latter principal symbol maps $\operatorname{ran}\widehat q$ bijectively onto the whole target fiber. Indeed the original weighted trace vector is the full Cauchy vector of the decaying normal solution, and $\mathcal B$ performs exactly the boundary measurements. The scalar positive order reductions are fiberwise isomorphisms at every nonzero covector.

Let $\widehat S,\widehat S''$ be the inverses just constructed, and undo (BF11). Then $$S_{kj}\in\Psi^{k-m_j}_{\rm cl}(Y;G_j,E),\qquad S''_{kl}\in\Psi^{k-l}_{\rm cl}(Y;E,E), \tag{BF12}$$ with $$\mathcal BS\equiv I,\quad QS\equiv S,\quad S\mathcal B+S''\equiv I,\quad S''Q\equiv0. \tag{BF13}$$ Consequently $S:\mathcal D^s\to\mathcal C^s$ and $S'':\mathcal C^s\to\mathcal C^s$ are continuous for every real $s$. For example the $(k,j)$ entry raises $s-m_j-1/2$ to $s-k-1/2$, since its order is $k-m_j$. This arithmetic is valid even when $m_j\geq m$, and does not replace the separate restriction $s\geq m$ needed for the realization.

5. A quantitative one-sided layer estimate

This item proves the analytic point needed even when a transmission operator is not an elliptic inverse. Let $T$ be a proper classical operator of integer order $-m$, $m\geq1$, having ordinary transmission across $t=0$, including every base and frequency jet. If $0\leq j<m$, define $K_jv=r^+T(v\otimes D_t^j\delta_0)$. Then, locally with fixed compact supports and globally after patching, $$\|K_jv\|_{\bar H^s(X^\circ)} \leq C_s\|v\|_{H^{s+j+1/2-m}(Y)} \qquad(s\in\mathbb R). \tag{BF14}$$ The same estimate holds for bundle matrices. A smooth off-diagonal kernel contributes a smoothing operator and will be retained as such throughout the proof.

We give the parameter estimate underlying (BF14). In a left-symbol chart let $a(y,t,\eta,\tau)$ be the symbol of $T$, let $\lambda=\langle\eta\rangle$, and set $$k_j(y,t,\eta)=(2\pi)^{-1}\int e^{it\tau} a(y,t,\eta,\tau)\tau^j\,d\tau,\quad t>0. \tag{BF15}$$ This is an oscillatory integral. It is not interpreted as an absolutely convergent integral after arbitrary differentiation. For all nonnegative integers $A,N$ and multi-indices $\alpha,\beta$, its one-sided extension satisfies $$\big|t^N\partial_t^A\partial_\eta^\alpha\partial_y^\beta k_j(y,t,\eta)\big| \leq C\lambda^{j+1-m+A-N-|\alpha|} \quad(t\geq0). \tag{BF16}$$ Equivalently its normal profile, after dividing by $\lambda^{j+1-m}$ and setting $z=\lambda t$, is bounded in every Schwartz seminorm on the closed half-line, with the usual loss $\lambda^{-|\alpha|}$ for tangential frequency derivatives. Constants involve only finitely many classical symbol and transmission seminorms for each displayed estimate.

Here are details of the normal-frequency argument. First freeze all normal base derivatives at zero. Write $d=j-m$, so initially $d\leq-1$, and rescale $\tau=\lambda\sigma$. On $|\sigma|\leq2$, differentiated symbols have the bounds of a compactly supported smooth function, times $\lambda^d$. On the two tails, Taylor expansion of each homogeneous component about the two normal rays gives integer powers of $\sigma$. Transmission makes their coefficients the same Laurent coefficients on the positive and negative real tails. More precisely, for any chosen number of tangential derivatives and any integer $L$, subtracting sufficiently many homogeneous components and sufficiently many normal-ray Taylor terms leaves a remainder bounded with those derivatives by $C\lambda^d\langle\sigma\rangle^{-L}$. If a tangential derivative is taken, include its extra factor $\lambda^{-1}$. This assertion follows directly from Taylor’s integral remainder on the compact normal-direction charts and the classical symbol remainder; choosing both truncation lengths larger than $L+|d|$ supplies the bound.

For any finite set of these Laurent coefficients, there is a rational function having exactly those coefficients: use the polynomial part and a finite linear combination of $(\sigma-i)^{-h}$, $h\geq1$. Expanding $(\sigma-i)^{-h}$ at infinity gives a triangular system with leading term $\sigma^{-h}$; solve it successively to the chosen order. Its inverse Fourier transform is a sum of derivatives of $\delta_0$ and, on $z>0$, exponential polynomials $c_hz^{h-1}e^{-z}$. Every such polynomial is smooth and rapidly decreasing on the closed positive half-line. The delta derivatives disappear after restriction to $z>0$; they are not incorrectly assigned point values at zero. The remainder can be made to have as many integrable weighted derivatives as desired, so its inverse Fourier transform and the desired $z$-derivatives and weights are bounded. The same subtraction applied after multiplying by powers of $\sigma$ justifies every normal derivative. This proves the frozen version of (BF16), including its limits at zero, by ordinary absolutely convergent integrals after subtraction. It is the quantitative use of all the transmission jets, not merely of the leading parity.

For the actual symbol expand in the normal base variable to length $L$:

$$a(y,t,\eta,\tau)=\sum_{l<L}t^l\partial_t^la(y,0,\eta,\tau)/l!+t^La_L(y,t,\eta,\tau)$$

Each frozen term has the estimate just proved. The remainder has the original symbol order, locally uniformly in $t$. To bound its inverse integral on $0<z=\lambda t\leq1$, divide the $\sigma$-axis into dyadic annuli. An annulus of radius $R$ contributes at most $C R^{d+A+1}$ before integration by parts, and after $M$ integrations contributes at most this bound times $(zR)^{-M}$. Splitting the sum at $R=z^{-1}$ gives $Cz^{-K}$, for an integer $K$ determined by the order and the requested derivatives. A logarithmic borderline sum is bounded by increasing $K$ by one. This exponent does not depend on $L$. Choose $L>K$, also allowing for the finitely many derivatives that hit $t^L$. Its factor $t^L=\lambda^{-L}z^L$ then removes the singularity. For $z\geq1$, integrate more often in $\sigma$; this gives any prescribed inverse power of $z$, and absorbs the polynomial $z^L$. Derivatives of the remainder in $t,y,\eta$ obey the same symbol bounds. Thus all finite sets of estimates (BF16) follow by choosing $L$ sufficiently large. This also proves smoothness up to zero without assuming that the symbol at a positive normal base point itself has matching normal-ray tails.

We next turn this profile estimate into (BF14). Extend a smooth half-line profile to the full line with bounded norms through any prescribed finite order. One explicit finite-order construction on the negative side is $h_-(z)=\chi(z)\sum_{l=1}^{N+1}c_l h(-lz)$, where the Vandermonde equations $\sum_lc_l(-l)^a=1$, $0\leq a\leq N$, match the first $N$ derivatives at zero. Away from zero use a cutoff and the already rapidly decreasing profile. The resulting extension is bounded in the finitely many weighted Sobolev norms in question; taking $N$ larger than their orders suffices. Apply this after normal rescaling, rather than using a fixed-scale extension on each frequency. It preserves the estimates in (BF16).

For completeness, the tangential operator bound needed here follows from the ordinary Fourier-kernel proof also for these normal profiles as a Hilbert-valued symbol. Localize to input tangential frequencies $|\eta|\asymp2^l$ and output frequencies $|\theta|\asymp2^h$. Fourier transformation in $y$ of the compactly supported symbol, with $M$ integrations by parts, gives the factor $\langle\theta-\eta\rangle^{-M}$. For $|h-l|\leq2$, change normal variable to $z=2^lt$. The $L^2_t$ factor is $2^{-l/2}$; each normal or tangential derivative costs at most $2^l$ by (BF16). Schur’s kernel bound and Plancherel therefore give, for any integer $N\geq0$, $$\|\Delta_h\widetilde K_j\Delta_l v\|_{H^N(\mathbb R^n)} \leq C_N 2^{l(N+j+1/2-m)}\|\Delta_l v\|_{L^2}, \quad |h-l|\leq2. \tag{BF17}$$ Here $\widetilde K_j$ is the chosen extension of the potential, and the low-frequency blocks use $2^l\geq1$. The estimate for a real nonnegative $s\leq N$ follows by weighted Fourier interpolation between the $N$ and zero estimates. For negative $s$, in a fixed output tangential annulus one has $(1+|\theta|^2+\tau^2)^{s/2}\leq C2^{hs}$, so the zero estimate gives the same exponent. When $|h-l|>2$, the factor from the $y$-integrations is bounded by any prescribed power of $2^{-\max(h,l)}$, after compensating the finite frequency volumes. The profile estimates remain valid after those derivatives. Thus the corresponding bound has an extra summable factor $2^{-M|h-l|}$, with $M$ arbitrarily large and with any fixed Sobolev powers absorbed by increasing $M$.

The dyadic characterization of Fourier Sobolev norms and the elementary convolution inequality for the summable sequence $2^{-M|h-l|}$ now sum (BF17). This gives a whole-space extension with norm at most the right side of (BF14), hence also the restriction norm. Density extends the map to every indicated boundary Sobolev space. Local trivializations and a finite partition of unity prove the bundle and compact-manifold versions. This completes the all-real estimate, including the negative exponents; interpolation from nonnegative integers alone would not have done so.

6. Interior forcing through a general transmission operator

For the same general transmission $T$ of order $-m$, $$\|r^+Te^+f\|_{\bar H^{m+r}(X^\circ)} \leq C_r\|f\|_{\bar H^r(X^\circ)}\qquad(r\geq0). \tag{BF18}$$ We prove this without claiming that $e^+:\bar H^r\to H^r$ is bounded for all $r$. At $r=0$, zero extension is an $L^2$ isometry, and whole-space pseudodifferential continuity proves (BF18).

Induct on the nonnegative integer $r$, simultaneously for all symbols $T$ with a bound controlled by finitely many seminorms. A tangential derivative commutes with $e^+$. For a normal derivative, distributional differentiation gives $$D_t(r^+Te^+f)=r^+Te^+D_tf+r^+[D_t,T]e^+f +i^{-1}r^+T((\gamma_0 f)\otimes\delta_0). \tag{BF19}$$ First prove this for smooth $f$, by differentiating its zero extension, and then use density. The commutator has order $-m$ and ordinary transmission, because its symbol is a normal base derivative of that of $T$. The induction estimate at $r-1$ bounds the first term in $H^{m+r-1}$; the second is bounded there from $f\in H^{r-1}$. The trace theorem gives $\gamma_0f\in H^{r-1/2}(Y)$, and (BF14) with $j=0$, $s=m+r-1$, bounds the last term in the same space. Tangential differentiation has the first two terms only. Since the integer Sobolev norm of order $m+r$ is controlled by the order-zero norm and these first-derivative norms of order $m+r-1$, the induction closes. Cutoff commutators are handled by the same symbol estimates; smooth off-diagonal kernels are harmless. All constants use finitely many seminorms, so the simultaneous induction is valid.

Interpolation of the restriction Sobolev scale between consecutive integers gives (BF18) for every real $r\geq0$. In particular the endpoint $r=0$ is included. This proof, together with Section 5, proves the two transmission estimates for arbitrary proper classical transmission operators of integer order $-m$. The operator need not satisfy $PT=I+R_F$. Its use in an actual boundary inverse will require that additional identity.

7. The boundary source and its anisotropic norm

For the specific $P^c$ in (BF4), normal coefficients must act on the delta derivatives before they are evaluated. The identity $fD_t^r\delta=\sum_{j=0}^r\binom rj(-1)^{r-j}(D_t^{r-j}f)(0)D_t^j\delta$, obtained by testing against a smooth function and applying Leibniz’s rule, gives the exact grouping $$\begin{split} P^cU&=\sum_{j=0}^{m-1}v_j\otimes D_t^j\delta_0,\\ v_j&=i^{-1}\sum_{\substack{l\geq0\\j+l+1\leq a\leq m}} \binom{a-1-l}{j}(-1)^{a-1-l-j} (D_t^{a-1-l-j}P_a)(y,0,D_y)U_l. \end{split} \tag{BF20}$$ When the coefficients are independent of the normal variable, only $a=j+l+1$ remains, giving $v_j=i^{-1}\sum_{l+j<m}P_{j+l+1}U_l$. For variable coefficients the additional normal derivatives in (BF20) are retained. This is precisely the same distributional source as (BF4), with no coefficient jet removed.

Each contribution from $U_l$ to the coefficient of $D_t^j\delta$ has tangential order at most $m-j-l-1$. Hence $\|v_j\|_{H^{s-m+j+1/2}}\leq C\sum_l\|U_l\|_{H^{s-l-1/2}}$. Using (BF14) in each term gives the full Poisson estimate $$\|KU\|_{\bar H^s(X^\circ)} \leq C_s\sum_{l=0}^{m-1}\|U_l\|_{H^{s-l-1/2}(Y)}, \qquad s\in\mathbb R. \tag{BF21}$$ This is also valid for a general transmission $T$ in place of the parametrix in (BF4), with the same differential boundary source.

We verify separately the anisotropic estimate that exposes the half-order and the restriction $j<m$. In a chart of dimension $n=\dim X$, define $$\|w\|_{a,b}^2=(2\pi)^{-n}\int (1+|\eta|^2+\tau^2)^a(1+|\eta|^2)^b |\widehat w(\eta,\tau)|^2\,d\eta\,d\tau. \tag{BF22}$$ For a boundary layer, $\widehat{v\otimes D_t^j\delta}=\widehat v(\eta)\tau^j$. Substitution $\tau=\langle\eta\rangle\sigma$ gives exactly $$\int_{\mathbb R}\tau^{2j}(1+|\eta|^2+\tau^2)^{-m}d\tau =c_{jm}\langle\eta\rangle^{2j-2m+1},\quad c_{jm}=\int_{\mathbb R}\sigma^{2j}(1+\sigma^2)^{-m}d\sigma<\infty. \tag{BF23}$$ At infinity the exponent is $2j-2m\leq-2$, and at zero it is nonnegative. Thus Editorial restoration of the Fourier factors. Comparing the $n$-dimensional measure in (BF22) with the $(n-1)$-dimensional boundary Sobolev measure gives the exact identity $$\|v\otimes D_t^j\delta\|_{-m,s}^2 =\frac{c_{jm}}{2\pi}\|v\|_{H^{s+j-m+1/2}}^2, \qquad c'_{jm}=\left(\frac{c_{jm}}{2\pi}\right)^{1/2}. \tag{BF23a}$$ This also holds in dimension one, using measure one on $\mathbb R^0$. The integral $c_{jm}$ remains exactly the one in (BF23). If $k$ is an integer with $k\geq\max(s,0)$, then $$\|w\|_{s-m-k,k}\leq\|w\|_{-m,s}, \tag{BF24}$$ because $(1+|\eta|^2+\tau^2)^{s-k}\leq\langle\eta\rangle^{2(s-k)}$. Commuting each of the finitely many tangential derivatives of order at most $k$ through a proper $T\in\Psi^{-m}$, and using whole-space Sobolev continuity on each resulting commutator, proves $T:H^{a-m,k}\to H^{a,k}$. The integer $k\geq0$ norm is equivalent to the sum of the $H^a$ norms of those derivatives, as is immediate from its Fourier weight. Therefore $$\|KU\|_{\bar H^{s-k,k}}\leq C_s \sum_l\|U_l\|_{H^{s-l-1/2}}. \tag{BF25}$$ This derivation holds at negative as well as positive $s$. For a parametrix, $PKU=r^+R_FP^cU$ is smooth, and elliptic normal recovery gives another route from (BF25) to (BF21). Our proof of (BF21) already supplied that recovery through the transmission profiles, so it does not use an unproved normal-regularity assertion as an extra assumption.